Chemistry Volume?

Question

What volume of a .250 molar solution of silver nitrate is required to react with 18.6 g of potassium phosphate? One of the products is potassium nitrate.

Answer 1

18.6g K3PO4 /212.2 g/mol = 0.0877mol
.0877mol K3PO4 x 3mol AgNO3/1mol K3PO4 =0.263 mol AgNO3
0.263 mol AgNO3 x 1 Liter/0.250 mol = 1.051 L = 1.05L (sig fig)

1051 mL or 1050 mL (3 sig figs)

Answer 2

The reaction is

K3PO4 + 3AgNO3 ===> Ag3PO4 + 3KNO3

From the stoichiometery, set up

3 x 18.6 / 212.3 = V x 0.25 /1000 where V is the volume of 0.25 M AgNO3 required in mls and 212.3 is the MW of K3PO4.

Solving leads to V= 1051 mls